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I'm in high school chemistry and just learning about buffer solutions. For example, given a buffered solution of $\ce{CH3COOH}$ and $\ce{NaCH3COO}$ with $K_a = 1.8 \times 10^{-5}$, we might want to find what the pH will become if $\ce{0.010 M KOH}$ is added. Given the dissociation of acetic acid:

$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$

My question is, which acid ($\ce{CH3COOH}$ or the conjugate $\ce{H3O+}$) will the base $\ce{KOH}$ react with? My teacher said that the base will react with $\ce{H3O+}$ to shift equilibrium in the forward direction, but why wouldn't it react with $\ce{CH3COOH}$ since it also is an acid?

Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?

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  • $\begingroup$ They're all in equilibrium together. It would be mistake to ask which one reacted. Whatever you pick is still part of the equilibrium that needs to readjust. $\endgroup$ – Zhe Apr 1 at 19:41
  • $\begingroup$ "My teacher said that the base will react with $\ce{H3O+}$ to shift equilibrium in the $\color{red}{reverse}$ direction,".I think to shift the equilibrium in the $\color{green}{forward}$ direction. $\endgroup$ – Adnan AL-Amleh Apr 1 at 20:34
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$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$

My question is, which acid (CH3COOH or the conjugate H3O+) will the base KOH react with?

You have to satisfy two equilibria, the one you wrote and the auto-dissociation of water.

$$\ce{H3O+ + OH- <=> 2 H2O}$$

If you add up the two reaction, you get a third one:

$$\ce{CH3COOH + OH- <=> H2O + CH3COO-}$$

You can go either way (use the second or the third reaction) to explain that $\ce{OH-}$ is part of a neutralization reaction. As a consequence, the concentration of $\ce{CH3COO-}$ increases and that of $\ce{CH3COOH}$ and hydronium decreases.

Hydroxide reacts with hydronium

In this case, you would invoke the second reaction. There is more hydroxide (you are adding sufficient KOH to raise the concentration to 0.01 M) than hydronium (at about pH = 5, the buffer will contain about $\pu{e-5}$ M). As the hydronium gets neutralized, the weak acid equilibrium (first reaction) will shift forward (not reverse - typo in the question), making more hydronium, which will get neutralized etc.

Hydroxide reacts with acetic acid

In a different explanation that results in the same answer, you could say the hydroxide turns some of the weak acid into weak base (third reaction). As a result, the weak acid / weak base concentration ratio will change, causing a net reverse shift in the weak acid equilibrium (first reaction) and lowering the hydronium concentration.

Which way is better?

It is personal preference. If we knew how fast the two reactions go, we might choose the faster one, but we don't. For figuring out the equilibrium concentrations and the pH, it does not matter.

Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?

No, and you can always combine two equilibrium reactions with overlapping species (like the hydronium ion in the first and second reaction) to arrive at a third that also has to be at equilibrium.

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  • $\begingroup$ "that the concentration of $~\ce{ OH−}$ $\color{red}{\text{drops}}$ and the concentration of $~\ce{ CH3COOH}$ $\color{red}{\text{increases}}$."I think:that the concentration of $\ce{ OH−}$ $\color{green}{\text{increases}}$ and the concentration of$\ce{ CH3COOH}$ $\color{green}{\text{decreases}}$. $\endgroup$ – Adnan AL-Amleh Apr 1 at 20:59
  • $\begingroup$ Thanks for the comprehensive answer. One question: you said that the direction I choose shouldn't affect pH. But in the Henderson-Hasselbalch equation don't you have to compute $-log([base]/[acid])$? If I choose different directions, won't the new concentrations of base and acid be different? $\endgroup$ – Andrew Li Apr 1 at 21:44
  • $\begingroup$ @Adnan AL-Amleh I fixed my typo: The neutralization reaction results in a drop of hydroxide and weak acid, while the concentration of weak base increases. $\endgroup$ – Karsten Theis Apr 1 at 21:55
  • $\begingroup$ @Andrew_Li I had a typo which I fixed, and I revised my answer to address your follow-up question. No matter which reaction you start with, the ratio of weak base to weak acid concentration will increase, so the pH will increase. You expect that because that is what strong bases do. Because the solution is buffered, the change is much smaller than it would be for pure water, but it still changes a bit in that direction. $\endgroup$ – Karsten Theis Apr 1 at 22:01
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The hydroxide ion reacts with the hydronium ion (I hate that name; I prefer the simpler "hydrogen ion") to increase the concentration of water and decrease the concentration of hydrogen ion. To restore the equilibrium, now more acetic acid will ionize, shifting the equilibrium to the right as you have it written above.

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Zhe's comment is not wrong, but I'd say that your teacher isn't either.

Consider a 0.1 molar solution of acetic acid. The water will be about 55 molar.

The twist here is that the proton of the hydronium ion is very liable in water. In other words, the proton exchange between water molecules happens very very fast. That is why using a mineral acid makes a solution very conductive. So just based on chance, a hydroxide ion is more likely to reaction with a hydronium ion than an acetic acid molecule. Hence your teacher's remark.

However Zhe is correct in that it really doesn't make a difference overall. The H+/acetate/acetic acid equilibrium will be established "fairly" quickly.

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